交叉冲印滤镜又叫做正片负冲滤镜，关于它的原理我一直没找到，希望有大佬给我甩个链接。
我用的测试图像在这里
我实现的交叉冲印是使用上图这样的点运算来实现的，下面是我的代码
void punchFilter(Mat src) {
int width = src.rows;
int height = src.cols;
int gray, value;
int rMap[256], gMap[256], bMap[256];
for (int i = 0; i < 256; i++){
value = i < 128 ? i : 256 - i;
gray = (int)pow(value, 3) / 16384; // value的3次方再除以16384
rMap[i] = i < 128 ? gray : 256 - gray;
gray = (int)pow(value, 2) / 128; // value的2次方再除以128
gMap[i] = i < 128 ? gray : 256 - gray;
bMap[i] = i / 2 + 0x25;//0x25 = 37
}
for (int j = 0; j < width - 1; j++) {
uchar* data = src.ptr<uchar>(j);
for (int i = 0; i < height - 1; i++) {
int b = bMap[data[i * 3]];
int g = gMap[data[i * 3 + 1]];
int r = rMap[data[i * 3 + 2]];
data[i * 3] = min(255, max(0, b));
data[i * 3 + 1] = min(255, max(0, g));
data[i * 3 + 2] = min(255, max(0, r));
}
}
imshow("交叉冲印滤镜", src);
}
效果图是这样的
这是怀旧风格滤镜，对原像素点的R、G、B进行变换
void recall(Mat src) {
int width = src.rows;
int height = src.cols;
Mat img(src.size(), CV_8UC3);
int x, y;
for (y = 0; y < height;y++) {
uchar* P0 = src.ptr<uchar>(y);
uchar* P1 = img.ptr<uchar>(y);
for (x = 0; x < width; x++) {
float R = P0[3 * x + 2];
float G = P0[3 * x + 1];
float B = P0[3 * x];
float newR = 0.393 * R + 0.769 * G + 0.189 * B;
float newG = 0.349 * R + 0.686 * G + 0.168 * B;
float newB = 0.272 * R + 0.534 * G + 0.131 * B;
if (newR < 0) newR = 0;
if (newR > 255) newR = 255;
if (newG < 0) newG = 0;
if (newG > 255) newG = 255;
if (newB < 0) newB = 0;
if (newB > 255) newB = 255;
P1[3 * x] = (uchar)newB;
P1[3 * x + 1] = (uchar)newG;
P1[3 * x + 2] = (uchar)newR;
}
}
imshow("怀旧滤镜实现", img);
}
效果图：
好像有点问题啊，图片不发黄啊
实现代码：
void light(int X, int Y, int K, int R, Mat src) {
int width = src.rows;
int height = src.cols;
double distance;
int x, y;
Mat img;
vector<Mat> m(3);
split(src, m);
for (y = 0; y < width ; y++) {
for (x = 0; x < height; x++) {
distance = sqrt((X - x) * (X - x) + (Y - y) * (Y - y));
if (distance < R) {
int temp = K * (0 > 1 - distance / R ? 0 : 1 - distance / R);
if (temp > 255) temp = 255;
int a = m[0].at<uchar>(x, y) + temp;
m[0].at<uchar>(x, y) = MIN(255, MAX(0, a));
a = m[1].at<uchar>(x, y) + temp;
m[1].at<uchar>(x, y) = MIN(255, MAX(0, a));
a = m[2].at<uchar>(x, y) + temp;
m[2].at<uchar>(x, y) = MIN(255, MAX(0, a));
}
}
}
merge(m, img);
imshow("光照滤镜实现", img);
}
效果图：
代码如下：
void leak(Mat src1, Mat src)
{
int width = src1.cols, height = src1.rows;
float newB;
float newG;
float newR;
Mat img(src1.size(), CV_8UC3);
for (int y = 0; y < height; y++)
{
uchar* P0 = src.ptr<uchar>(y);
uchar* P1 = src1.ptr<uchar>(y);
uchar* P3 = img.ptr<uchar>(y);
for (int x = 0; x < width; x++)
{
if (P1[3 * x] <= 128)
{
newB = P0[3 * x] * P1[3 * x] / 128;
}
else
newB = 255 - ((255 - P0[3 * x]) * (255 - P1[3 * x])) / 128;
if (newB > 255)
{
newB = 255;
}
if (P1[3 * x + 1] <= 128)
{
newG = P0[3 * x + 1] * P1[3 * x + 1] / 128;
}
else
newG = 255 - ((255 - P0[3 * x + 1]) * (255 - P1[3 * x + 1])) / 128;
if (newG > 255)
{
newG = 255;
}
if (P1[3 * x + 2] <= 128)
{
newR = P0[3 * x + 2] * P1[3 * x + 2] / 128;
}
else
newR = 255 - ((255 - P0[3 * x + 2]) * (255 - P1[3 * x + 2])) / 128;
if (newR > 255)
{
newR = 255;
}
P3[3 * x] = newB;
P3[3 * x + 1] = newG;
P3[3 * x + 2] = newR;
}imshow("漏光滤镜", img);
}
}
漏光模板，用PS随便拉个前景色就好了
其实自己写的部分也没有多少,大部分还是在网上拼拼凑凑,终于得到了我想要的结果而已.
经过课设发现自己对于数字图像这块还是学习还是不够的,趁着寒假努力点吧.
最后附上整个项目的代码:
#include <opencv2/opencv.hpp>
#include "opencv2/highgui/highgui.hpp"
#include "opencv2/imgproc/imgproc.hpp"
#include <stdio.h>
#include <iostream>
using namespace cv;
using namespace std;
//交叉冲印
void punchFilter(Mat src);
//怀旧风格滤镜
void recall(Mat src);
//光照效果滤镜
void light(int X, int Y, int K, int R, Mat src);//XY是光源坐标，K光照系数，R光照半径
//漏光滤镜
void leak(Mat src, Mat msak);
int main(int argc, char** argv) {
while (1) {
cout << "请选择显示的滤镜效果" << endl;
cout << "1、交叉冲印滤镜" << endl;
cout << "2、怀旧风格滤镜" << endl;
cout << "3、光照效果滤镜" << endl;
cout << "4、漏光滤镜" << endl;
int choice;
cin >> choice;
switch (choice) {
case 1: {
Mat src = imread(argv[1]);
Mat p_temp(src.rows, src.cols, CV_8UC3, src.data);
imshow("原图", src);
punchFilter(p_temp);
break;
}
case 2: {
Mat src = imread(argv[1]);
Mat r_temp(src.rows, src.cols, CV_8UC3, src.data);
imshow("原图", src);
recall(r_temp);
break;
}
case 3: {
Mat src = imread(argv[1]);
Mat l_temp(src.rows, src.cols, CV_8UC3, src.data);
int X, Y, K, R;
cout << "请输入光源坐标(x,y),光照系数，光照半径" << endl;
cin >> X >> Y >> K >> R;
imshow("原图", src);
light(X, Y, K, R, l_temp);
break;
}
case 4: {
Mat src = imread(argv[1]);
Mat leak_temp(src.rows, src.cols, CV_8UC3, src.data);
Mat mask = imread("C:\\Users\\18704\\Desktop\\mask.png");
imshow("原图", src);
leak(leak_temp, mask);
}
default:{
cout << "无效字符，请重新输入" << endl;
break;
}
}
waitKey(0);
}
return 0;
}
void punchFilter(Mat src) {
int width = src.rows;
int height = src.cols;
int gray, value;
int rMap[256], gMap[256], bMap[256];
for (int i = 0; i < 256; i++){
value = i < 128 ? i : 256 - i;
gray = (int)pow(value, 3) / 16384; // value的3次方再除以16384
rMap[i] = i < 128 ? gray : 256 - gray;
gray = (int)pow(value, 2) / 128; // value的2次方再除以128
gMap[i] = i < 128 ? gray : 256 - gray;
bMap[i] = i / 2 + 0x25;//0x25 = 37
}
for (int j = 0; j < width - 1; j++) {
uchar* data = src.ptr<uchar>(j);
for (int i = 0; i < height - 1; i++) {
int b = bMap[data[i * 3]];
int g = gMap[data[i * 3 + 1]];
int r = rMap[data[i * 3 + 2]];
data[i * 3] = min(255, max(0, b));
data[i * 3 + 1] = min(255, max(0, g));
data[i * 3 + 2] = min(255, max(0, r));
}
}
imshow("交叉冲印滤镜", src);
}
void recall(Mat src) {
int width = src.rows;
int height = src.cols;
Mat img(src.size(), CV_8UC3);
int x, y;
for (y = 0; y < height;y++) {
uchar* P0 = src.ptr<uchar>(y);
uchar* P1 = img.ptr<uchar>(y);
for (x = 0; x < width; x++) {
float R = P0[3 * x + 2];
float G = P0[3 * x + 1];
float B = P0[3 * x];
float newR = 0.393 * R + 0.769 * G + 0.189 * B;
float newG = 0.349 * R + 0.686 * G + 0.168 * B;
float newB = 0.272 * R + 0.534 * G + 0.131 * B;
if (newR < 0) newR = 0;
if (newR > 255) newR = 255;
if (newG < 0) newG = 0;
if (newG > 255) newG = 255;
if (newB < 0) newB = 0;
if (newB > 255) newB = 255;
P1[3 * x] = (uchar)newB;
P1[3 * x + 1] = (uchar)newG;
P1[3 * x + 2] = (uchar)newR;
}
}
imshow("怀旧滤镜实现", img);
}
void light(int X, int Y, int K, int R, Mat src) {
int width = src.rows;
int height = src.cols;
double distance;
int x, y;
Mat img;
vector<Mat> m(3);
split(src, m);
for (y = 0; y < width ; y++) {
for (x = 0; x < height; x++) {
distance = sqrt((X - x) * (X - x) + (Y - y) * (Y - y));
if (distance < R) {
int temp = K * (0 > 1 - distance / R ? 0 : 1 - distance / R);
if (temp > 255) temp = 255;
int a = m[0].at<uchar>(x, y) + temp;
m[0].at<uchar>(x, y) = MIN(255, MAX(0, a));
a = m[1].at<uchar>(x, y) + temp;
m[1].at<uchar>(x, y) = MIN(255, MAX(0, a));
a = m[2].at<uchar>(x, y) + temp;
m[2].at<uchar>(x, y) = MIN(255, MAX(0, a));
}
}
}
merge(m, img);
imshow("光照滤镜实现", img);
}
void leak(Mat src1, Mat src)
{
int width = src1.cols, height = src1.rows;
float newB;
float newG;
float newR;
Mat img(src1.size(), CV_8UC3);
for (int y = 0; y < height; y++)
{
uchar* P0 = src.ptr<uchar>(y);
uchar* P1 = src1.ptr<uchar>(y);
uchar* P3 = img.ptr<uchar>(y);
for (int x = 0; x < width; x++)
{
if (P1[3 * x] <= 128)
{
newB = P0[3 * x] * P1[3 * x] / 128;
}
else
newB = 255 - ((255 - P0[3 * x]) * (255 - P1[3 * x])) / 128;
if (newB > 255)
{
newB = 255;
}
if (P1[3 * x + 1] <= 128)
{
newG = P0[3 * x + 1] * P1[3 * x + 1] / 128;
}
else
newG = 255 - ((255 - P0[3 * x + 1]) * (255 - P1[3 * x + 1])) / 128;
if (newG > 255)
{
newG = 255;
}
if (P1[3 * x + 2] <= 128)
{
newR = P0[3 * x + 2] * P1[3 * x + 2] / 128;
}
else
newR = 255 - ((255 - P0[3 * x + 2]) * (255 - P1[3 * x + 2])) / 128;
if (newR > 255)
{
newR = 255;
}
P3[3 * x] = newB;
P3[3 * x + 1] = newG;
P3[3 * x + 2] = newR;
}imshow("漏光滤镜", img);
}
}
来源：https://www.cnblogs.com/lumping-boy/p/12164045.html